During the seventh-inning stretch of a baseball game, groundskeepers drag mats a
ID: 1960403 • Letter: D
Question
During the seventh-inning stretch of a baseball game, groundskeepers drag mats across the infield dirt to smooth it. A groundskeeper is pulling a mat at a constant velocity by applying a force of 120 N at an angle of 22degree above the horizontal. The coefficient of kinetic friction between the mat and the ground is 0.60. Find (a) the magnitude of the frictional force between the dirt and the mat and (b) the weight of the mat.***************
We have the answer to this question in the text book but do not know how it was achieved. How can one solve for frictional force without knowing the mass?
Explanation / Answer
Given that Applied force F = 120 N at an angle = 22 deg Coefficient of friction = 0.6 (a) By applying Newton's laws along the horizontal F cos - f = ma Since when it is moving with constant velocity Fcos -f = 0 Therefore the frictional force f = Fcos = (120 N) cos22 = 111.26 N -------------------------------------------------------------------------- (b) Now by applying the Newton's laws along vertical Fsin + N = mg Therefore the normal force N = mg - Fsin Now the frictional force f =N 111.26 N = (0.6) [mg - Fsin] 111.26 N = (0.6)mg - (0.6)(120 N) sin22 111.26 N = (0.6)mg - (0.6)(120 N) sin22 111.26 + 26.97 = (0.6) mg Therefore the weight of the mat is mg = 230.38 N 111.26 N = (0.6) [mg - Fsin] 111.26 N = (0.6)mg - (0.6)(120 N) sin22 111.26 N = (0.6)mg - (0.6)(120 N) sin22 111.26 + 26.97 = (0.6) mg Therefore the weight of the mat is mg = 230.38 NRelated Questions
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